Tap chi Toan Hong kong 15.2

Volume 15, Number 2 July - September, 2010 Lagrange Interpolation Formula Kin Y. Li Olympiad Corner Below are the problems used in the selection of the Indian team for IMO-2010. Problem 1. Is there a positive integer n, which is a multiple of 103, such that 2 2n+1 ≡2 (mod n)? Problem 2. Let a, b, c be integers such that b is even. Suppose the equation x 3 +ax 2 +bx+c=0 has roots α, β, γ such that α 2 = β+γ. Prove that α is an integer and β≠γ. Problem 3. Let ABC be a triangle in which BC < AC. Let M be the midpoint of AB; AP be the altitude from A on to BC; and BQ be the altitude from B on to AC. Suppose QP produced meet AB (extended) in T. If H is the orthocenter of ABC, prove that TH is perpendicular to CM. Problem 4. Let ABCD be a cyclic quadrilateral and let E be the point of intersection of its diagonals AC and BD. Suppose AD and BC meet in F. Let the midpoints of AB and CD be G and H respectively. If Γ is the circumcircle of triangle EGH, prove that FE is tangent to Γ. (continued on page 4) Editors: 張 百 康 (CHEUNG Pak-Hong), Munsang College, HK 高 子 眉 (KO Tsz-Mei) 梁 達 榮 (LEUNG Tat-Wing) 李 健 賢 (LI Kin-Yin), Dept. of Math., HKUST 吳 鏡 波 (NG Keng-Po Roger), ITC, HKPU Artist: 楊 秀 英 (YEUNG Sau-Ying Camille), MFA, CU Acknowledgment: Thanks to Elina Chiu, Math. Dept., HKUST for general assistance. On-line: http://www.math.ust.hk/mathematical_excalibur/ The editors welcome contributions from all teachers and students. With your submission, please include your name, address, school, email, telephone and fax numbers (if available). Electronic submissions, especially in MS Word, are encouraged. The deadline for receiving material for the next issue is October 20, 2010. For individual subscription for the next five issues for the 10-11 academic year, send us five stamped self-addressed envelopes. Send all correspondence to: Dr. Kin-Yin LI, Math Dept., Hong Kong Univ. of Science and Technology, Clear Water Bay, Kowloon, Hong Kong Fax: (852) 2358 1643 Email: makyli@ust.hk © Department of Mathematics, The Hong Kong University of Science and Technology Let n be a positive integer. If we are given two collections of n+1 real (or complex) numbers w 0 , w 1 , …, w n and c 0 , c 1 , …, c n with the w k ’s distinct, then there exists a unique polynomial P(x) of degree at most n satisfying P(w k ) = c k for k = 0,1,…,n. The uniqueness is clear since if Q(x) is also such a polynomial, then P(x)−Q(x) would be a polynomial of degree at most n and have roots at the n+1 numbers w 0 , w 1 , …, w n , which leads to P(x)−Q(x) be the zero polynomial. Now, to exhibit such a polynomial, we define f 0 (x)=(x−w 1 )(x−w 2 )⋯(x−w n ) and similarly for i from 1 to n, define f i (x)=(x−w 0 )⋯(x−w i−1 )(x−w i+1 )⋯(x−w n ). Observe that f i (w k ) = 0 if and only if i≠k. Using this, we see ∑ = = n i ii i i wf xf cxP 0 )( )( )( satisfies P(w k ) = c k for k = 0,1,…,n. This is the famous Lagrange interpolation formula. Below we will present some examples of using this formula to solve math problems. Example 1. (Romanian Proposal to 1981 IMO) Let P be a polynomial of degree n satisfying for k = 0,1,…,n, . 1 )( 1− ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = k n kP Determine P(n+1). Solution. For k = 0,1,…,n, let w k =k and . )!1( )!1(! 1 1 + −+ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = − n knk k n c k Define f 0 , f 1 , … , f n as above. We get f k (k) = (−1) n−k k!(n−k)! an d . )1( )!1( )1( kn n nf k −+ + =+ By the Lagrange interpolation formula, ,)1( )( )1( )1( 00 ∑∑ == − −= + =+ n k n k kn k k k kf nf cnP which is 0 if n is odd and 1 if n is even. Example 2. (Vietnamese Proposal to 1977 IMO) Suppose x 0 , x 1 , …, x n are integers and x 0 > x 1 > ⋯ > x n . Prove that one of the numbers |P(x 0 )|, |P(x 1 )|, … , |P(x n )| is at least n!/2 n , where P(x) = x n + a 1 x n–1 + ⋯ + a n is a polynomial with real coefficients. Solution. Define f 0 , f 1 , … , f n using x 0 , x 1 , …, x n . By the Lagrange interpolation formula, we have , )( )( )()( 0 ∑ = = n i ii i i xf xf xPxP since both sides are polynomials of degrees at most n and are equal at x 0 , x 1 , …, x n . Comparing coefficients of x n , we get ∑ = = n i ii i xf xP 0 . )( )( 1 Since x 0 , x 1 , …, x n are strictly decreasing integers, we have ∏∏ += − = −−= n ij ij i j ijii xxxxxf 1 1 0 |||||)(| . ! 1 )!(! ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =−≥ i n n ini Let the maximum of |P(x 0 )|, |P(x 1 )|, … , |P(x n )| be |P(x k )|. By the triangle inequality, we have . ! )(2 ! )( )(| )( 1 00 n xP i n n xP xf xP k n n i k n i ii i = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≤≤ ∑∑ == Then |P(x k )| ≥ n!/2 n . Example 3. Let P be a point on the plane of ∆ABC. Prove that .3≥++ AB PC CA PB BC PA Mathematical Excalibur, Vol. 15, No. 2, Jul. - Sep. 2010 Page 2 Solution. We may take the plane of ∆ABC to be the complex plane and let P, A, B, C be corresponded to the complex numbers w, w 1 , w 2 , w 3 respectively. Then PA =|w–w 1 |, BC=|w 2 –w 3 |, etc. Now the only polynomial P(x) of degree at most 2 that equals 1 at w 1 , w 2 , w 3 is the constant polynomial P(x) ≡ 1. So, expressing P(x) by the Lagrange interpolation formula, we have ))(( ))(( ))(( ))(( 3121 32 2313 21 wwww wxwx wwww wxwx −− −− + −− −− .1 ))(( ))(( 1232 13 ≡ −− −− + wwww wxwx Next, setting x = w and applying the triangle inequality, we get .1≥++ BC PA AB PC AB PC CA PB CA PB BC PA (*) The inequality (r+s+t) 2 ≥ 3(rs+st+tw), after subtracting the two sides, reduces to [(r–s) 2 +(s–t) 2 +(t–r) 2 ]/2 ≥ 0, which is true. Setting r= PA/BC, s=PB/CA and t=PC/AB, we get .3 2 ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++≥ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ++ BC PA AB PC AB PC CA PB CA PB BC PA AB PC CA PB BC PA Taking square roots of both sides and applying (*), we get the desired inequality. Example 4. (2002 USAMO) Prove that any monic polynomial (a polynomial with leading coefficient 1) of degree n with real coefficients is the average of two monic polynomials of degree n with n real roots. Solution. Suppose F(x) is a monic real polynomial. Choose real y 1 , y 2 , … ,y n such that for odd i, y i < min{0,2F(i)} and for even i, y i > max{0,2F(i)}. By the Lagrange interpolation formula, there is a polynomial of degree less than n such that P(i) = y i for i=1,2,…,n. Let G(x) = P(x)+(x−1)(x−2)⋯(x−n) and H(x) = 2F(x)−G(x). Then G(x) and H(x) are monic real polynomials of degree n and their average is F(x). As y 1 , y 3 , y 5 , … < 0 and y 2 , y 4 , y 6 , ⋯ > 0, G(i)=y i and G(i+1)=y i+1 have opposite signs (hence G(x) has a root in [i,i+1]) for i=1,2,…,n−1. So G(x) has at least n−1 real roots. The other root must also be real since non-real roots come in conjugate pair. Therefore, all roots of G(x) are real. Similarly, for odd i, G(i) = y i < 2F(i) implies H(i)=2F(i)−G(i) > 0 and for even i, G(i) = y i > 2F(i) implies H(i) = 2F(i)−G(i) < 0. These imply H(x) has n real roots by reasoning similar to G(x). Example 5. Let a 1 , a 2 , a 3 , a 4 , b 1 , b 2 , b 3 , b 4 be real numbers such that b i –a j ≠0 for i,j=1,2,3,4. Suppose there is a unique set of numbers X 1 , X 2 , X 3 , X 4 such that ,1 41 4 31 3 21 2 11 1 = − + − + − + − ab X ab X ab X ab X ,1 42 4 32 3 22 2 12 1 = − + − + − + − ab X ab X ab X ab X ,1 43 4 33 3 23 2 13 1 = − + − + − + − ab X ab X ab X ab X .1 44 4 34 3 24 2 14 1 = − + − + − + − ab X ab X ab X ab X Determine X 1 +X 2 +X 3 +X 4 in terms of the a i ’s and b i ’s. Solution. Let .)()()( 4 1 4 1 ∏∏ == −−−= ii ii bxaxxP Then the coefficient of x 3 in P(x) is . 4 1 4 1 ∑∑ == − ii ii ab Define f 1 , f 2 , f 3 , f 4 using a 1 , a 2 , a 3 , a 4 as above to get the Lagrange interpolation formula . )( )( )()( 4 ∑ = = ii ii i i af xf aPxP Since the coefficient of x 3 in f i (x) is 1, the coefficient of x 3 in P(x) is also . )( )( 4 1 ∑ =i ii i af aP Next, observe that P(b j )/f i (b j ) = b j – a i , which are the denominators of the four given equations! For j = 1,2,3,4, setting x = b j in the interpolation formula and dividing both sides by P(b j ), we get . )(/)( )( )( )( )( 1 44 1 ∑∑ == − == iii ij iii ii ji j i ab afaP af bf bP aP Comparing with the given equations, by uniqueness, we get X i =P(a i )/f i (a i ) for i = 1,2,3,4. So . )( )( 4 1 44 1 4 1 ∑∑∑∑ ==== −== ii ii i ii i i i ab af aP X Comment: This example is inspired by problem 15 of the 1984 American Invitational Mathematics Examination. Example 6. (Italian Proposal to 1997 IMO) Let p be a prime number and let P(x) be a polynomial of degree d with integer coefficients such that: (i) P(0) = 0, P(1) = 1; (ii) for every positive integer n, the remainder of the division of P(n) by p is either 0 or 1. Prove that d ≥ p − 1. Solution. By (i) and (ii), we see P(0)+P(1)+ ⋯+P(p − 1)≡k (mod p) (#) for some k ∈{1,2,…, p − 1}. Assume d ≤ p − 2. Then P(x) will be uniquely determined by the values P(0), P(1), …, P(p − 2). Define f 0 , f 1 , …, f p−2 using 0, 1, …, p − 2 as above to get the Lagrange interpolation formula . )( )( )()( 2 0 ∑ − = = p k k k kf xf kPxP As in example (1), we have f k (k) = (−1) p−2−k k!(p−2−k)!, kp p pf k −− − =− 1 )!1( )1( and so . 1 )1)(()1( 2 0 ∑ − = − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − −=− p k kp k p kPpP Next, we claim that .20)(mod)1( 1 −≤≤−≡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − pkforp k p k This is true for k = 0. Now for 0 < i < p, )(mod0 )!(! ! p ipi p i p ≡ − = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ because p divides p!, but not i!(p−i)!. If the claim is true for k, then )(mod)1( 1 11 1 1 p k p k p k p k + −≡ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + − and the induction step follows. Finally the claim yields ∑ − = −≡− 2 0 ).(mod)()1()1( p k p pkPpP So P(0)+P(1)+ ⋯+P(p − 1)≡ 0 (mod p), a contradiction to (#) above. Mathematical Excalibur, Vol. 15, No. 2, Jul. - Sep. 2010 Page 3 Problem Corner We welcome readers to submit their solutions to the problems posed below for publication consideration. The solutions should be preceded by the solver’s name, home (or email) address and school affiliation. Please send submissions to Dr. Kin Y. Li, Department of Mathematics, The Hong Kong University of Science & Technology, Clear Water Bay, Kowloon, Hong Kong. The deadline for sending solutions is October 20, 2010. Problem 351. Let S be a unit sphere with center O. Can there be three arcs on S such that each is a 300 ° arc on some circle with O as center and no two of the arcs intersect? Problem 352. (Proposed by Pedro Henrique O. PANTOJA, University of Lisbon, Portugal) Let a, b, c be real numbers that are at least 1. Prove that . 2 3 111 222 ≥ + + + + + ab abc ca cab bc bca Problem 353. Determine all pairs (x, y) of integers such that x 5 −y 2 =4. Problem 354. For 20 boxers, find the least number n such that there exists a schedule of n matches between pairs of them so that for every three boxers, two of them will face each other in one of the matches. Problem 355. In a plane, there are two similar convex quadrilaterals ABCD and AB 1 C 1 D 1 such that C, D are inside AB 1 C 1 D 1 and B is outside AB 1 C 1 D 1 Prove that if lines BB 1 , CC 1 and DD 1 concur, then ABCD is cyclic. Is the converse also true? ***************** Solutions **************** Problem 346. Let k be a positive integer. Divide 3k pebbles into five piles (with possibly unequal number of pebbles). Operate on the five piles by selecting three of them and removing one pebble from each of the three piles. If it is possible to remove all pebbles after k operations, then we say it is a harmonious ending. Determine a necessary and sufficient condition for a harmonious ending to exist in terms of the number k and the distribution of pebbles in the five piles. (Source: 2008 Zhejiang Province High School Math Competition) Solution. CHOW Tseung Man (True Light Girl’ s College), CHUNG Ping Ngai (MIT Year 1), HUNG Ka Kin Kenneth (CalTech Year 1). The necessary and sufficient condition is every pile has at most k pebbles in the beginning. The necessity is clear. If there is a pile with more than k pebbles in the beginning, then in each of the k operations, we can only remove at most 1 pebble from that pile, hence we cannot empty the pile after k operations. For the sufficiency, we will prove by induction. In the case k=1, three pebbles are distributed with each pebble to a different pile. So we can finish in one operation. Suppose the cases less than k are true. For case k, since 3k pebbles are distributed. So at most 3 piles have k pebbles. In the first operation, we remove one pebble from each of the three piles with the maximum numbers of pebbles. This will take us to a case less than k. We are done by the inductive assumption. Problem 347. P(x) is a polynomial of degree n such that for all w ∈{1, 2, 2 2 , …, 2 n }, we have P(w) = 1/w. Determine P(0) with proof. Solution 1. Carlo PAGANO (Università di Roma “Tor Vergata”, Roma, Italy). William CHAN Wai-lam (Carmel Alison Lam Foundation Secondary School) and Thien Nguyen (Nguyen Van Thien Luong High School, Dong Nai Province, Vietnam). Let Q(x) = xP(x)−1 = a(x−1)(x−2) ⋯(x−2 n ). For x ≠1, 2, 2 2 , …, 2 n , . 2 1 2 1 1 1 )( )(' n xxxxQ xQ − ++ − + − = L Since Q(0)= −1 and Q’(x)=P(x)+xP’(x), ∑ = −==−== n k nk Q Q QP 0 . 2 1 2 2 1 )0( )0(' )0(')0( Solution 2. CHUNG Ping Ngai (MIT Year 1), HUNG Ka Kin Kenneth (CalTech Year 1), Abby LEE (SKH Lam Woo Memorial Secondary School, Form 5) and WONG Kam Wing (HKUST, Physics, Year 2). Let Q(x) = xP(x)−1 = a(x−1)(x−2) ⋯(x−2 n ). Now Q(0) = −1 = a(−1) n+1 2 s , where s = 1+2+ ⋯+n. So a = (−1) n 2 −s . Then P(0) is the coefficient of x in Q(x), which is ∑ = −− −==+++− n k nk nsssn a 0 1 . 2 1 2 2 1 )222()1( L Other commended solvers: Samuel Lilό ABDALLA (ITA-UNESP, São Paulo, Brazil), Problem 348. In ∆ ABC, we have ∠BAC = 90° and AB < AC. Let D be the foot of the perpendicular from A to side BC. Let I 1 and I 2 be the incenters of ∆ ABD and ∆ ACD respectively. The circumcircle of ∆ AI 1 I 2 (with center O) intersects sides AB and AC at E and F respectively. Let M be the intersection of lines EF and BC. Prove that I 1 or I 2 is the incenter of the ∆ ODM, while the other one is an excenter of ∆ ODM. (Source: 2008 Jiangxi Province Math Competition) Solution. CHOW Tseung Man (True Light Girl’ s College). A B C D I 1 I 2 O F E M We claim EF intersects AD at O. Since ∠EAF=90°, EF is a diameter through O. Next we will show O is on AD. Since AI 1 , AI 2 bisect ∠BAD, ∠CAD respectively, we get ∠I 1 AI 2 =45°. Then ∠I 1 OI 2 =90°. Since OI 1 =OI 2 , ∠OI 1 I 2 =45°. Also, DI 1 , DI 2 bisect ∠BDA, ∠CDA respectively implies ∠I 1 DI 2 =90°. Then D, I 1 , O, I 2 are concyclic. So .45 2212 ADIIOIODI ∠==∠=∠ o Then O is on AD and the claim is true. Since ∠EOI 1 = 2∠EAI 1 = 2∠DAI 1 = ∠DOI 1 and I 1 is on the angle bisector of ∠ODM, we see I 1 is the incenter of ∆ ODM. Similarly, replacing E by F and I 1 by I 2 in the last sentence, we see I 2 is an excenter of ∆ ODM. Other commended solvers: CHUNG Ping Ngai (MIT Year 1), HUNG Ka Kin Kenneth (CalTech Year 1) and Abby LEE (SKH Lam Woo Memorial Secondary School, Form 5). Problem 349. Let a 1 , a 2 , …, a n be rational numbers such that for every positive integer m, m n mm aaa +++ L 21 is an integer. Prove that a 1 , a 2 , …, a n are integers. Mathematical Excalibur, Vol. 15, No. 2, Jul. - Sep. 2010 Page 4 Solution. CHUNG Ping Ngai (MIT Year 1) and HUNG Ka Kin Kenneth (CalTech Year 1). We may first remove all the integers among a 1 , a 2 , …, a n since their m-th powers are integers, so the rest of a 1 , a 2 , …, a n will still have the same property. Hence, without loss of generality, we may assume all a 1 , a 2 , …, a n are rational numbers and not integers. First write every a i in simplest term. Let Q be their least common denominator and for all 1≤i≤n, let a i =k i /Q. Take a prime factor p of Q. Then p is not a prime factor of one of the k i ’s. So one of the remainders r i when k i is divided by p is nonzero! Since k i ≡r i (mod p), so for every positive integer m, ).(mod0 111 mm n i m i n i m i n i m i pQakr ≡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =≡ ∑∑∑ === This implies . 1 ∑ = ≤ n i m i m rp Since r i < p, ,0lim 1 lim1 11 = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =≤ ∑∑ = ∞→ = ∞→ n i m i m n i m i m m p r r p which is a contradiction. Comments: In the above solution, it does not need all positive integers m, just an infinite sequence of positive integers m with the given property will be sufficient. Problem 350. Prove that there exists a positive constant c such that for all positive integer n and all real numbers a 1 , a 2 , …, a n , if P(x) = (x − a 1 )(x − a 2 ) ⋯ (x − a n ), then .)(max)(max ]1,0[]2,0[ xPcxP x n x ∈∈ ≤ (Ed Both solutions below show the conclusion holds for any polynomial!) Solution 1. LEE Kai Seng. Let S be the maximum of |P(x)| for all x ∈[0,1]. For i=0,1,2,…,n, let b i =i/n and ).())(()()( 110 niii bxbxbxbxxf −−−−= +− LL By the Lagrange interpolation formula, for all real x, . )( )( )()( 0 ∑ = = n i ii i i bf xf bPxP For every w ∈[0,2], |w−b k | ≤ |2−b k | for all k = 0,1,2,…,n. So ∏ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=≤ n i ii n i fwf 0 2)2()( n n nnnn )1()22)(12(2 + − − = L . ! )!2( n nn n = Also, |P(b i )| ≤ S and . )!(! )( n ii n ini bf − = By the triangle inequality, . 22 )( )( )()( 00 ∑∑ == ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≤≤ n i ii i n i i n in i n S bf wf bPwP Finally, .2 2 2 2222 0 42 0 ∑∑ == ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ ≤ ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ n i nn n i n n n n i n n in i n Then .)(max162)(max ]1,0[ 4 ]2,0[ xPSwP x nn w ∈∈ =≤ Solution 2. G.R.A.20 Problem Solving Group (Roma, Italy). For a bounded closed interval I and polynomial f(x), let ||f|| I denote the maximum of |f(x)| for all x in I. The Chebyschev polynomial of order n is defined by T 0 (x) = 1, T 1 (x) = x and T n (x) = 2xT n−1 (x)−T n−2 (x) for n ≥ 2. (Ed By induction , we can obtain T n (x) = 2 n x n +c n−1 x n−1 + ⋯ + c 0 and T n (cos θ)=cos nθ. So T n (cos(πk/n))= (−1) k , which implies all n roots of T n (x) are in (−1,1) as it changes sign n times.) It is known that for any polynomial Q(x) with degree at most n>0 and all t ∉[−1,1], |Q(t)| ≤ ||Q|| [−1,1] |T n (t)|. (!) To see this, we may assume ||Q|| [−1,1] = 1 by dividing Q(x) by such maximum. Assume x 0 ∉[−1,1] and |Q(x 0 )| > |T n (x 0 )|. Let a = T(x 0 )/Q(x 0 ) and R(x) = aQ(x)−T n (x). For k = 0, 1, 2, ⋯, n, since T n (cos(πk/n)) = (−1) k and |a|<1, we see R(cos(πk/n)) is positive or negative depending on whether k is odd or even. (In particular, R(x) ≢0.) By continuity, R(x) has n+1 distinct roots on [−1,1] ∪{x 0 }, which contradicts the degree of R(x) is at most n. Next, for the problem, we claim that for every t ∈[1,2], we have |P(t)| ≤ 6 n ||P|| [0,1] . (Ed Observe that the change of variable t = (s+1)/2 is a bijection between s ∈[−1,1] and t∈[0,1]. It is also a bijection between s ∈[1,3] and t∈[1,2].) By letting Q(s) = P((s+1)/2), the claim is equivalent to proving that for every s ∈[1,3], we have |Q(s)|≤ 6 n ||Q|| [−1,1] . By (!) above, it suffices to show that |T n (s)| ≤ 6 n for every s∈[1,3]. Clearly, |T 0 (s)|=1=6 0 . For n=1 and s ∈[1,3], |T 1 (s)|=s≤3<6. Next, since the largest root of T n is less than 1, we see all T n (s) > 0 for all s∈[1,3]. Suppose cases n−2 and n−1 are true. Then for all s ∈[1,3], we have 2sT n−1 (s), T n−2 (s) > 0 and so |T n (s)| = |2sT n−1 (s)−T n−2 (s)| ≤ max(2sT n−1 (s), T n−2 (s)) ≤ max(6·6 n−1 , 6 n−2 ) = 6 n . This finishes everything. Olympiad Corner (continued from page 1) Problem 5. Let A=(a jk ) be a 10×10 array of positive real numbers such that the sum of the numbers in each row as well as in each column is 1. Show that there exist j<k and l<m such that . 50 1 ≥+ kljmkmjl aaaa Problem 6. Let ABC be a triangle. Let AD, BE, CF be cevians such that ∠BAD=∠CBE=∠ACF. Suppose these cevians concur at a point Ω. (Such a point exists for each triangle and it is called a Brocard point.) Prove that .1 2 2 2 2 2 2 ≥ Ω + Ω + Ω AB C CA B B C A (Ed A cevian is a line segment which joins a vertex of a triangle to a point on the opposite side or its extension.) Problem 7. Find all functions f:ℝ→ℝ such that f(x+y) + xy = f(x)f(y) for all reals x,y. Problem 8. Prove that there are infinitely many positive integers m for which there exist consecutive odd positive integers p m , q m (=p m +2) such that the pairs (p m , q m ) are all distinct and 2222 , mmmmmmmm qqmppqqpp ++++ are both perfect squares. . a(x−1)(x 2) ⋯(x 2 n ). For x ≠1, 2, 2 2 , …, 2 n , . 2 1 2 1 1 1 )( )(' n xxxxQ xQ − ++ − + − = L Since Q(0)= −1 and Q’(x)=P(x)+xP’(x), ∑ = −==−== n k nk Q Q QP 0 . 2 1 2 2 1 )0( )0(' )0(')0( . ,1 41 4 31 3 21 2 11 1 = − + − + − + − ab X ab X ab X ab X ,1 42 4 32 3 22 2 12 1 = − + − + − + − ab X ab X ab X ab X ,1 43 4 33 3 23 2 13 1 = − + − + − + − ab X ab X ab X ab X .1 44 4 34 3 24 2 14 1 = − + − + − + − ab X ab X ab X ab X . . )( )( )()( 0 ∑ = = n i ii i i bf xf bPxP For every w ∈[0 ,2] , |w−b k | ≤ |2 b k | for all k = 0,1 ,2, …,n. So ∏ = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=≤ n i ii n i fwf 0 2) 2()( n n nnnn )1( )22 )( 12( 2 + − − = L . ! ) !2( n nn n = Also, |P(b i )|